7th Maths : Term 3 Unit 3 : Algebra : Inequations : Text Book Back Exercises Questions with Answers, Solution

**Exercise
3.2**

**1.
Given that x ≥ y. Fill in the blanks with suitable inequality signs.**

**Answer: **

**(i) y ≤ x
**

**(ii) x + 6 ≥ y + 6**

**(iii) x²≥ xy
**

**(iv) – xy ≤ **

**(v) ) x – y **

** **

**2.
Say True or False.**

(i)
Linear inequation has almost one solution. **[Answer: False]**

(ii)
When *x* is an integer, the solutions for *x* ≤ 0 are −1, −2,... **[Answer: False]**

(iii)
An inequation, −3 < *x* < −1, where *x* is an integer, cannot be
represented in the number line. **[Answer: True]**

*(iv)
x *< −*y *can be rewritten as −*y *<* x *

** **

**3.
Solve the following inequations.**

*(i)
x ***≤ 7, where x is a natural number.**

*(ii)
x ***− 6 < 1, where x is a natural number.**

**(iii)
2 a + 3 ≤ 13, where a is a whole number.**

**(iv)
6 x − 7 ≥ 35, where x is an integer.**

**(v)
4 x − 9 > −33, where x is a negative integer.**

**Solution: **

**(i)** *x* ≤ 7, where *x* is a
natural number.

Since the solution belongs to the set of natural numbers, that
are less than or equal to 7, we take the values of *x *as 1,2, 3, 4, 5, 6 and 7.

**(ii)** **x – 6 < 1, where x is
a natural number.**

x – 6 < 1 Adding 6 on the both the sides *x* – 6 + 6 < 1 + 6

x < 7

Since the solutions belongs to the set of natural numbers that
are less than 7, we take the values of *x*
as 1,2, 3, 4, 5 and 6

**(iii) 2 a + 3 ≤ 13, where a is a whole number.**

2*a* + 3 ≤ 13

Subtracting 3 from both the sides

2*a* + 3 – 3 ≤ 13 – 3

2*a *≤ 10

Dividing both the sides by 2.

2*a*/2 ≤ 10/2

a ≤ 5

Since the solutions belongs to the set of whole numbers that are
less than or equal to 5 we take the values of *a* as 0, 1, 2, 3, 4 and 5

**(iv)** **6 x – 7 ≥ 35, where x is an integer.**

6*x* – 7 **≥** 35 Adding 7 on both the sides

6*x* –7 + 7 **≥ **35 + 7

6*x ***≥** 42

Dividing both the sides by 6 we get

6*x*/6 **≥ **42/6

x **≥ **7

Since the solution belongs to the set of integers that are
greater than or equal to 7, we take the values of *x* as 7, 8, 9, 10...

**(v) 4 x – 9 > –33, where x is a negative integer.**

4*x* – 9 > – 33 Adding
9 both the sides

4*x* – 9 + 9 > –33 +
9

4*x* > – 24

Dividing both the sides by 4

4*x*/4 > –24/4

*x* > – 6

Since the solution belongs to a negative integer that are
greater than –6, we take values of *u*
as –5, –4, –3, –2 and–1.

** **

**4.
Solve the following inequations and represent the solution on the number line:**

*(i)
k ***> −5, k is an integer.**

Since the solution belongs
to the set of integers, the solution is –4, –3, –2, –1, 0,... It’s graph on number
line is shown below.

**(ii)
−7 ≤ y, y is a negative integer.**

–7 ≤ *y*

Since the solution set belongs to the set of negative integers,
the solution is –7,–6,–5,–4,–3,–2,–1.

Its graph on the number line is shown below

**(iii)
−4 ≤ x ≤ 8, x is a natural number**

–4 ≤ x ≤ 8

Since the solution belongs to the set of natural numbers, the
solution is 1,2, 3,4, 5, 6, 7 and 8.

Its graph on number line is shown below

**(iv)
3 m − 5 ≤ 2m + 1, m is an integer.**

3*m* – 5 ≤ 2*m* + 1

Subtracting 1 on both the sides

3*m* – 5 – 1 ≤ 2*m* + 1 + 1

3*m* – 6 ≤ 2*m*

Subtracting 2m on both the sides

3*m* – 6 – 2*m* ≤ 2*m*
– 2*m*

*m* – 6 ≤ 0

Adding 6 on both the sides

*m* – 6 + 6 ≤ 0 + 6

*m* ≤ 6

Since the solution belongs to the set of integers, the solution
is 6, 5, 4, 3, 2, 1, 0, –1,...

Its graph on number line is shown below

** **

**5.
An artist can spend any amount between ₹**** ****80 to ₹**** ****200
on brushes. If cost of each brush is ₹**** ****5 and there are 6 brushes in each packet,
then how many packets of brush can the artist buy?**

**Solution:**

Given the artist can spend any amount between ₹ 80 to ₹ 200

Let the number of packets of brush he can buy be *x*

Given cost of 1 brush = ₹ 5

Cost of 1 packet brush (6 brushes) = ₹5 x 6 = ₹30

∴ Cost of *x* packets of
brushes = 30*x*

∴ The inequation becomes 80 ≤ 30*x *≤ 200

Dividing throughout by 30 we get 80/30 ≤ 30*x*/30 ≤ 200/30

8/3 ≤ *x *≤
20/3 ; 2 (2/3) ≤ *x *≤ 6 (2/3)

brush packets cannot get in fractions.

∴ The artist can buy 3 ≤ *x*
≤ 6 packets of brushes.

or *x* = 3, 4, 5 and 6
packets of brushes.

** **

__Objective Type Questions__

**6.
The solutions of the inequation 3 ≤ p ≤ 6 are (where p is a natural
number)**

(i) 4, 5 and 6

(ii) 3, 4 and 5

(iii) 4 and 5

(iv) 3, 4, 5 and 6

**[Answer: (iv) 3,4,5 and 6]**

**7.
The solution of the inequation 5 x + 5 ≤ 15 are (where x is a natural
number)**

(i) 1 and 2

(ii) 0, 1 and 2

(iii) 2, 1, 0, −1, −2..

(iv) 1, 2, 3..

**[Answer: (i) 1 and
2]**

**Solution: **5*x* + 5 ≤ 15

5*x* < 15 – 5 = 10

*x *≤ 10/5 = 2

**8.
The cost of one pen is ₹**** ****8 and it is available in a sealed pack
of 10 pens. If Swetha has only ₹**** ****500, how many packs of pens can she buy
at the maximum?**

(i) 10

(ii) 5

(iii) 6

(iv) 8

**[Answer: (iii) 6]**

**Solution: **Price of 1 pen = ₹8

Price of 1 pack = 10 × 8 = 80

Number of packs Swetha can buy =* x*

80*x* ≤ 500

8*x* ≤ 50

*x* ≤ = 50/8 = 6.25

*x* is a natural number *x*
= 1, 2, 3, 4, 5, 6

**9.
The inequation that is represented on the number line as shown below is _______**

(i) -4 <* x *< 0

(ii) −4 ≤* x *≤ 0

(iii) −4 <* x *≤ 0

(iv) −4 ≤* x *< 0

**[Answer: * –4 ≤ x
≤ 2]**

__ANSWERS:__

**Exercise 3.2**

1. (i)* y *≤*
x *(ii)* x *+ 6 ≥* y *+ 6 (iii) x^{2} ≥ xy (iv)
–xy≤–y^{2} (v) x–y ≥ 0

2. (i) False (ii)
False (iii) True (iv) False

3. (i)* x *= 1, 2, 3, 4, 5, 6 and 7

(ii)* x *= 1, 2, 3, 4, 5 and 6

(iii) *a* = 0, 1, 2, 3, 4 and 5

(iv)* x *=
7, 8, 9, 10, ....

(v)* x *=
–5, –4, –3, –2 and –1

4. (i) *k* = – 4, –3, –2, –1, 0, 1, 2, ...

(ii) *y* = – 7, –6, –5, –4, –3, –2 and –1

(iii) *x* = 1, 2, 3, 4, 5, 6, 7 and 8

(iv) *m* = ...–3, -2, –1, 0, 1, 2, 3, 4, 5, 6

5. The artist can buy
3 ≤ *x* ≤ 6 brushes or *x* = 3, 4, 5 and 6 brushes.

**Objective type questions**

6. (iv) 3, 4, 5 and 6

7. (i) 1 and 2

8. (iii) 6

9. (ii) –4 ≤* x *≤ 0 / –4 ≤* x *≤ 2

Tags : Questions with Answers, Solution | Algebra | Term 3 Chapter 3 | 7th Maths , 7th Maths : Term 3 Unit 3 : Algebra

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7th Maths : Term 3 Unit 3 : Algebra : Exercise 3.2 (Inequations) | Questions with Answers, Solution | Algebra | Term 3 Chapter 3 | 7th Maths

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